Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10010;
void open() {
  freopen("goal.in", "r", stdin);
  freopen("goal.out", "w", stdout);
}
class Gauss {
 public:
  int a[MAXN], b[MAXN], n;
  vector<pair<int, int> > ops[2];
  void work() {
    cin >> n;
    bool fail = 0;
    for (int i = 1; i <= n; ++i) scanf("%d", a + i);
    for (int i = 1; i <= n; ++i) scanf("%d", b + i);
    elimination(a, ops[0]);
    elimination(b, ops[1]);
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 0) break;
      int k = 1 << 30;
      while (!(a[i] & k)) k >>= 1;
      for (int j = 1; j <= min(n, 31); ++j)
        if ((a[j] ^ b[j]) & k) opera(a, j, i, ops[0]);
    }
    for (int i = 1; i <= n; ++i)
      if (a[i] != b[i]) fail = 1;
    if (fail)
      cout << -1 << endl;
    else {
      cout << ops[0].size() + ops[1].size() << endl;
      for (int i = 0; i < ops[0].size(); ++i)
        printf("%d %d\n", ops[0][i].first, ops[0][i].second);
      for (int i = ops[1].size() - 1; i >= 0; --i)
        printf("%d %d\n", ops[1][i].first, ops[1][i].second);
    }
  }

 private:
  void opera(int *x, int i, int j, vector<pair<int, int> > &g) {
    x[i] ^= x[j];
    g.push_back(make_pair(i, j));
  }
  void swap(int *x, int i, int j, vector<pair<int, int> > &g) {
    x[i] ^= x[j];
    x[j] ^= x[i];
    x[i] ^= x[j];
    g.push_back(make_pair(i, j));
    g.push_back(make_pair(j, i));
    g.push_back(make_pair(i, j));
  }
  void elimination(int *x, vector<pair<int, int> > &g) {
    int r = 1;
    for (int i = 30; i >= 0; --i) {
      for (int j = r; j <= n; ++j)
        if (x[j] & (1 << i)) {
          if (j != r) {
            swap(x, r, j, g);
          }
          for (int k = 1; k <= n; ++k) {
            if (k != r && (x[k] & (1 << i))) opera(x, k, r, g);
          }
          ++r;
          break;
        }
    }
  }
} newton;
int main() { newton.work(); }